# 26 is unique between a square (5^2) and a cube (3^3)

1. Oct 29, 2010

### al-mahed

i.e, $$a^{2} \pm 2 = b^{3}$$ has (25,27) as the only solution.

Now, can you prove it?

2. Oct 30, 2010

### robert Ihnot

I think this is difficult problem. However, there is a trivial solution (1)^2-2=(-1)^3.

3. Oct 30, 2010

### al-mahed

Hi Robert, I'm sorry I didn't clarify that a, b must be positive integers.

4. Oct 30, 2010

### Petek

Are you asking for help in proving the statement or challenging us to give a proof?

5. Oct 30, 2010

### al-mahed

second option :)

6. Oct 30, 2010

### hochs

It's easy if you use Z[sqrt(-2)] and Z[sqrt(2)] (they're both UFD).

For example look at a^2 + 2 = b^3. This factors as (a + sqrt(-2))(a - sqrt(-2)) = b^3, and the units of Z[sqrt(-2)] are just +/-1, so

a + sqrt(-2) = (c + d*sqrt(-2))^3 for some c,d in Z.

So

a = c^3 - 6cd^2 = c*(c^2 - 6d^2) and
1 = d*(3c^2 - 2d^2).

So d = +/-1 and c = +/- 1, whence you get a = +/-5, and so b = 2.

Same situation for a^2 - 2 = b^3.

7. Oct 30, 2010

### robert Ihnot

Seem O.K. to me, except b=3. In the second case we have a=+/-1 b=-1. However, Dr. Math makes it more complicated: http://mathforum.org/library/drmath/view/51569.html

8. Oct 31, 2010

### hochs

Yea, I don't know why the post in that link is using both a + sqrt(-2) = w^3 and a - sqrt(-2) = t^3 (conjugate). You only need the first, you get just as much information just more quickly.

9. Oct 31, 2010

### Petek

To go from

$$(a + \sqrt{-2})(a - \sqrt{-2}) = b^3$$

to

$$a + \sqrt{-2} = (c + d\sqrt{-2})^3$$

don't we need that $a + \sqrt{-2}$ and $a - \sqrt{-2}$ are relatively prime? In fact, as Doctor Rob points out on the Math Forum, GCD($a + \sqrt{-2}, a - \sqrt{-2}$) divides $2\sqrt{-2}$. I think that's why he has to consider several cases.

10. Oct 31, 2010

### hochs

Yes, that's right. So what I wrote above is not entirely correct, but one gets the idea - just factor and use UFD-ness of Z[sqrt(-2)].

More importantly, does anyone know how to use tex here without having to type itex and \itex everytime I want it? for example i prefer to , just much easier.

11. Oct 31, 2010

### Petek

The pdf file linked at the beginning of this thread strongly suggests that itex, /itex and its variants are the only format that works here.

12. Nov 2, 2010

### al-mahed

Hi folks, I followed a different direction.

first notice that a and b must have the same parity, so first case:

1) a, b = even

$$(2k)^{2}\pm 2 = (2n)^{3}$$
$$4k^{2}\pm 2 = 8n^{3}$$
so,
$$2k^{2}\pm 1= 4n^{3}$$

absurd, since even number cannot be equal to odd number, so second case:

2) a, b = odd

$$(2k+1)^{2}\pm 2 = (2n+1)^{3}$$

2.1) + 2

$$4k(k+1)+3 = 8n^{3}+12n^{2}+6n+1$$

wich means

$$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4$$
$$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k$$
and
$$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (K+1)$$

so

$$4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4$$
$$4n^{3}+6n^{2}+3n\equiv\ 1 \mod k$$
and
$$4n^{3}+6n^{2}+3n\equiv\ 1 \mod (K+1)$$

given $$f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .... + a_{i}x^{m_{i}}$$
and $$g(x) = b_{0} + b_{1}x + b_{2}x^{2} + b_{3}x^{3} + ....+ b_{j}x^{n_{j}}$$

we know that

$$f(x)\equiv\ g(x)\mod W$$ $$\Leftrightarrow$$
$$a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W$$

lets make

$$f(x) = 4n^{3}+6n^{2}+3n$$

and

$$g(x) = 1$$

finally (skip obvious manipulations)

$$12\equiv\ 0 \mod 4$$
$$12\equiv\ 0 \mod k$$
and
$$12\equiv\ 0 \mod (K+1)$$

k must be = 2

2.2) - 2

(...) (after the same steps)

finally

$$14\equiv\ 0 \mod 4$$

what is untrue

Any remarks would be apreciated.

13. Nov 2, 2010

### al-mahed

Is there anything not clear, or perhaps wrong? I apreciate some feedback

Last edited: Nov 2, 2010
14. Nov 4, 2010

### al-mahed

Thinking again, I suspect I misunderstood the theorem found here:

http://planetmath.org/encyclopedia/PolynomialCongruence.html

and jumped to a conclusion prematurely.

$$f(x)\equiv\ g(x)\mod W$$
$$a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W$$

Seems that the conclusion above is wrong.

Anyway, I'd preciate if anyone could make the theorem more clear to me, and make any comments about what I've posted :)

thank you!

15. Nov 4, 2010

### al-mahed

$$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4$$
$$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k$$
and
$$8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (k+1)$$

so

$$4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4$$
$$4n^{3}+6n^{2}+3n\equiv\ 1 \mod k$$
and
$$4n^{3}+6n^{2}+3n\equiv\ 1 \mod (k+1)$$

that's also wrong, but do not detroy the following argument because

$$24\equiv\ 0 \mod 4$$
$$24\equiv\ 0 \mod k$$
and
$$24\equiv\ 0 \mod (k+1)$$

still would imply that k must be = 2